Question

The Rolle’s theorem is applicable in the interval $-1\le x\le 1$ for the function

• A. $f\left(x\right)=x$
• B. $f\left(x\right)=x^2$
• C. $f\left(x\right)=2x^3+3$
• D. $f\left(x\right)=\left|x\right|$

Answer 1

The correct option is B

$f\left(x\right)=x^2$

Explanation for correct option:

Option(B): Rolle's theorem states that if a function $f$ is continuous on the closed interval $\left[a,b\right]$ and differentiable on the open interval $\left(a,b\right)$ such that $f\left(a\right)=f\left(b\right)$, then $f\text{'}\left(x\right)=0$ for some $x$ with $a\le x\le b$

Given interval is $-1\le x\le 1$

$f\left(x\right)=x^2$
$$\begin{gathered} \Rightarrow f(-1)=1 \\ \Rightarrow f\left(1\right)=1=f(-1) \end{gathered}$$
Function is continuous in given interval $\Rightarrow$ Rolle's theorem is applicable

Thus, option(B) is correct

Explanation for incorrect options:

O,ption(A): $f\left(x\right)=x$
$$\begin{gathered} \Rightarrow f(-1)=-1 \\ \Rightarrow f\left(1\right)=1\ne f(-1) \end{gathered}$$
$\Rightarrow$Rolle's theorem is not applicable

Thus, option(A) is incorrect
Option(C) : $f\left(x\right)=2x^3+3$
$$\begin{gathered} \Rightarrow f(-1)=-2+3 \\ =1 \\ \Rightarrow f\left(1\right)=2+3 \\ =5\ne f(-1) \end{gathered}$$
$\Rightarrow$Rolle's theorem is not applicable

Thus, option(C) is incorrect.

Option(D) : $f\left(x\right)=\left|x\right|$
$$\begin{gathered} \Rightarrow f(-1)=1 \\ \Rightarrow f\left(1\right)=1=f(-1) \end{gathered}$$

But the function is not differentiable because it has sharp point at $x=0$.

Function is continuous but not differentiable in given interval $\Rightarrow$Rolle's theorem is not applicable

Thus, option(D) is incorrect.
Hence option(B) is correct.